PROBLEM: $M\subset\mathbb{R}^n$ is a open subset, $p\in M$ is arbitrary. Find $T_pM$ and $N_pM$.
I know how to determine $T_pM$ and $N_pM$ for explicit examples, but I dont know how to handle this general problem.
What I tried so far:
$$f:\mathbb{R}^n\rightarrow\mathbb{R}$$ $$T_pM=\ker (Df(x_1,\dots,x_n)|_p)=\ker(\partial_1 f(p),\dots,\partial_nf(p))$$ $$=\{(x_1,\dots,x_n)\in\mathbb{R}^n:(x_1,\dots, x_n)\cdot(\partial_1 f(p),\dots,\partial_nf(p))=0\}$$ $$\Rightarrow x_1\partial_1 f(p)+\dots+x_n\partial_nf(p)=0 $$
The answer is that for an open set $M$ in $\mathbb R^n$, and $p\in M$, the tangent space $T_pM=\mathbb R^n$, and $N_pM=0$.
Here's the reason: If you know some differential topology, you will be clear that each tangent vector $v\in T_pM$ is a equivalent class of beam of osculating curves. To show that $T_pM=\mathbb R^n$, you need only to show that each equivalent class is nonempty, that is, to show for each nonzero vector $w\in \mathbb R^n$ , there is a curve $f_w:(-\varepsilon,+\varepsilon)\to M$ with $f_w(0)=p$, such that the tangent vector of the curve at point $p$ is $w$ itself. That is $f_w'(0)=w.$
Apparently there are many such functions. For example the linear function $f_w(x):=p+xw$, where $x\in (-\varepsilon,+\varepsilon)$. Now use the openess of $M$ to show that the for each nonzero vector $w\in \mathbb R^n$ , there exist such a curve $f_w$. You have completed the first part of proof.
As for the normal space $N_pM$, you may know that it is actually complement space of $T_pM$ in $\mathbb R^n$, so it is zero space.